# How do we know that wave f. is the eigenfunction of an operator H?

## Main Question or Discussion Point

I am kind of new to this eigenvalue, eigenfunction and operator things, but i have come across this quote many times:
##\psi## is the eigenfunction of an operator ##\hat{H}## with eigenvalue ##W##.
First i need some explaination on how do we know this? All i know about operator ##\hat{H}## so far is this equation where ##\langle W \rangle## is an energy expected value:

\begin{align}
\langle W \rangle &= \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x
\end{align}

From which it follows that ##\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p##.

I know how to derive relation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## for which they state that:

##\hat{a} \psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W-\hbar \omega)##.
I also know how to derive relation ##\hat{H}\hat{a}^\dagger = (W + \hbar \omega)\hat{a}^\dagger \psi## for which they state that:

##\hat{a}^\dagger \psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W+\hbar \omega)##.
How do we know this?

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HallsofIvy
Homework Helper
That's pretty much the definition of "eigenfunction" and "eigenvalue", isn't it?

That's pretty much the definition of "eigenfunction" and "eigenvalue", isn't it?
What exactly here is the definition? Which equation?

What exactly here is the definition? Which equation?
Any equation of the form
##\hat{H} \psi = E \psi##
is an eigen-equation

The functions ##\psi_n## which satisfy this equation are called the eigenfunctions of the operator ##\hat{H}##, with corresponding eigenvalues ##E_n##.

Put simply, when an operator ##\hat{H}## acts upon one of its eigenfunctions, it returns the same eigenfunction multiplied by a number, which we call the eigenvalue.

Any equation of the form
##\hat{H} \psi = E \psi##
is an eigen-equation

The functions ##\psi_n## which satisfy this equation are called the eigenfunctions of the operator ##\hat{H}##, with corresponding eigenvalues ##E_n##.

Put simply, when an operator ##\hat{H}## acts upon one of its eigenfunctions, it returns the same eigenfunction multiplied by a number, which we call the eigenvalue.
1st:
In this equation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## is it possible that ##\hat{a} \psi## is an eigenfunction... i mean ##\hat{a}## is an operator...

2nd:
And is it possible that my quotings were wrong i will write the wrong one and the one that i think is better:

WRONG:
##\hat{a}\psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##.

RIGHT:
##\hat{a}\psi## is an eigenfunction of operators ##\hat{H}\hat{a}## with eigenvalue ##(W−\hbar \omega)##.

Nugatory
Mentor
1st:
In this equation ##\hat{H}\hat{a} = (W - \hbar \omega)\hat{a} \psi## is it possible that ##\hat{a} \psi## is an eigenfunction... i mean ##\hat{a}## is an operator...

2nd:
And is it possible that my quotings were wrong i will write the wrong one and the one that i think is better:

WRONG:
##\hat{a}\psi## is an eigenfunction of operator ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##.

RIGHT:
##\hat{a}\psi## is an eigenfunction of operators ##\hat{H}\hat{a}## with eigenvalue ##(W−\hbar \omega)##.
What's right is wrong and what's wrong is right An operator maps one function into another, so if ##\hat{a}## is an operator and ##\psi## is a function, then ##\hat{a}\psi## is itself a function.

And I think that you dropped a ##\psi## from the equation you started with. I would have expected ##\hat{H}\hat{a}\psi = (W - \hbar \omega)\hat{a} \psi##, and written that way it is clear that the function ##\hat{a}\psi## is an eigenfunction of ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##... Or at least adding a few parentheses will make it clear:

$$\hat{H}(\hat{a}\psi) = (W - \hbar \omega)(\hat{a} \psi)$$

What's right is wrong and what's wrong is right An operator maps one function into another, so if ##\hat{a}## is an operator and ##\psi## is a function, then ##\hat{a}\psi## is itself a function.

And I think that you dropped a ##\psi## from the equation you started with. I would have expected ##\hat{H}\hat{a}\psi = (W - \hbar \omega)\hat{a} \psi##, and written that way it is clear that the function ##\hat{a}\psi## is an eigenfunction of ##\hat{H}## with eigenvalue ##(W−\hbar \omega)##... Or at least adding a few parentheses will make it clear:

$$\hat{H}(\hat{a}\psi) = (W - \hbar \omega)(\hat{a} \psi)$$
Ok so the eigenfunction can be a combination of an operator $\hat{a}$ and a function $\psi$?

Ok so the eigenfunction can be a combination of an operator $\hat{a}$ and a function $\psi$?
Why not? An operator acting on a function gives you another function.

Thanks.